Goodness of Fit Tests

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Goodness of Fit Tests

• QSCI 381 Lecture 40
• (Larson and Farber, Sect 10.1)

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Multinomial Experiments

• A
  is a probability experiment consisting of a
  fixed number of trials in which there are more
  than two possible outcomes for each independent
  trial. The probability for each outcome is fixed
  and each outcome is classified into
  
• .
• Examples of multinomial experiments include
• You sample 100 animals from a population. The
  categories could be age, length, maturity state.
• You sample 1000 poppies in a field. The
  categories could be colour.
• You sample 20 animals and calculate the frequency
  that each has a particular genetic haplotype.

multinomial experiment
categories
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Goodness-of-fit Tests

• A
  is used to test whether an observed
  frequency distribution fits an expected
  distribution.
• We need to specify a null and an alternative
  hypothesis. Generally the null hypothesis is that
  the observed frequency distribution (the data)
  fits the expected distribution. The alternative
  hypothesis is that this is not the case.

chi-square goodness-of-fit test
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Example-I

• We expect that a healthy marine mammal
  population should consist of an equal number of
  males and females, and that 60 of the population
  should be mature. We sample 150 animals and
  assess the fraction in each of four categories to
  be

Mature Female Mature Male Immature Female Immature Male
30 40 32 48
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Observed and Expected Frequencies

• The
  of a category is the frequency for the category
  observed in the data.
• The
  of a category is the calculated frequency for the
  category. Expected frequencies are obtained by
  assuming the specified (or hypothesized)
  distribution is correct. The expected frequency
  for the i th category is
• Where n is the number of trials, and pi is the
  assumed probability for the i th category.

observed frequency O
expected frequency E
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Observed and Expected Frequencies(Example)
Mature Female Mature Male Immature Female Immature Male
Observed frequency 30 40 32 48
Assumed probability 0.3 0.3 0.2 0.2
Expected frequency 45 (150 x 0.3) 45 (150 x 0.3) 30 (150 x 0.2) 30 (150 x 0.2)
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The Chi-square goodness-of-fit Test-I

• IF
• the observed frequencies are obtained from a
  random sample, and
• the expected frequencies are greater than or
  equal to 5 (pool categories if this is not the
  case).
• then the sampling distribution for the
  goodness-of-fit test is a chi-square distribution
  with k-1 degrees of freedom where k is the number
  of categories. The test statistic is

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The Chi-square goodness-of-fit Test-II

• Identify the claim and state the null and
  alternative hypotheses.
• Specify the level of significance, ?.
• Determine the degrees of freedom, d.fk-1.
• Find the critical value of the chi-square
  distribution and hence define the rejection
  region for the test.
• Calculate the test statistic.
• Check whether or not the value of the test
  statistic is in the rejection region.

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Example (Test using ?0.01)

• H0 the distribution of animals between sex and
  maturity classes equals that expected for a
  healthy population.
• The degrees of freedomk-13.
• The critical value of the chi-square distribution
  is 11.34 (CHIINV(0.01,3))

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Example (Test using ?0.01)
Mature Female Mature Male Immature Female Immature Male
Observed frequency 30 40 32 48
Expected frequency 45 45 30 30
5 0.56 0.13 10.80

• We reject the null hypothesis at the
• 1 level of significance.

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Example-A-1 (?0.05)

• The probability of a particular bird species
  utilizing each of five habitats is known. We
  collect data for a different species (n137) and
  wish to assess whether the two species differ in
  their habitat requirements.

Habitat type Habitat type Habitat type Habitat type Habitat type
1 2 3 4 5
Expected p 0.2 0.1 0.05 0.5 0.15
Observed 30 17 0 72 18
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Example-A-2 (?0.05)
Habitat type Habitat type Habitat type Habitat type Habitat type
1 2 3 4 5
Observed frequency 30 17 0 72 18
Expected frequency 27.4 13.7 6.85 68.5 20.55
0.25 0.79 6.85 0.18 0.32
The critical value is 9.49 we fail to reject
the null hypothesis
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Testing for Normality

• We can use the chi-square test in some cases to
  assess whether a variable is normally
  distributed.
• The null and alternative hypotheses are that
• The variable has a normal distribution.
• The variable does not have a normal distribution.

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Example
Class boundaries Frequency
5-15 6
15-25 23
25-35 53
35-45 45
45-55 22
Can we assume that these data are normal (assume
?0.05)?
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Calculating the Test Statistic
Class boundaries Observed frequency O Cumulative normal Cumulative normal Expected p Expected Frequency E
Lower Upper Difference
5-15 6 0.0030 0.0368 0.0338 5.0 0.1822
15-25 23 0.0368 0.2037 0.1669 24.9 0.1407
25-35 53 0.2037 0.5526 0.3488 52.0 0.0202
35-45 45 0.5526 0.8627 0.3102 46.2 0.0318
45-55 22 0.8637 0.9800 0.1172 17.5 1.1746
? 149 0.977 145.57 1.5497
Eipi x 149
xi is the mid-point of each class